# Question #6fd67

##### 1 Answer

#### Answer:

**(B)** *KInetic energy at P is equal to potential energy at Q*

#### Explanation:

The idea here is that the energy that the ball has in point **converted** to another type of energy as it reaches point

So, the ball starts at a point *potential energy* associated with the height at which point *above the ground*. You can thus say that

#E_P = m * g * p -># potential energy at p meteres above the ground

Now, what would the ball *need* in order to go from a point **higher** point

SInce it's being thrown, an **initial velocity** directed straight upward must be applied to the ball. This means that at point **kinetic component** as well.

Let's assume that the ball is being thrown with a velocity

#E_P = m * g * p + 1/2 * mv^2#

What will happen as the ball moves upward?

Since it's being acted upon by *gravity*, which always points **downward**, its initial velocity will *decrease* as the height of the ball increases.

This means that its *kinetic energy* will decrease as well. At its **highest point**, the ball's velocity will be equal to **zero**.

SInce the height at which the ball is located increases, its *potential energy* will increase as well. Let's assume that point **above the ground**.

In point

#E_Q = m * g * q + 1/2 * m * 0^2 = m * g * q#

Since the ball started from

Since the energy of the ball **must be conserved** in points

#E_P = E_Q#

#m * g * p + 1/2 * m * v^2 = m * g * q#

If you rearrange this equation, you'll get

#1/2 * m * v^2 = m * g * q - m * g * p#

#1/2 * m * v^2 = m * g * (q-p)#

This tells you that the **kinetic energy** that the ball had in point **potential energy** in point

If you take point

#1/2 * m * v^2 = m * g * q -># kinetic energy at p is equal to potential energy at q